Special Relativity.

The standard imaginary model of special relativity is that of a beam of light (AB) seen as perpendicular to the motion inside the reference frame. While the observer outside the moving reference frame sees the beam of light (AC) pointing ahead of the perpendicular.

If v is the speed of the moving reference frame and if t is the time inside the reference frame and t’ the time outside:
AB = ct, AC = ct’ and BC = vt’
As ABC is a right angle triangle:
(ct)² = (ct’)² - (vt’)²
t = t’ √(1 - v²/c²) ... (1)

Now imagine that the beam of light, as seen inside the reference frame (AB), is pointing slightly ahead of the perpendicular.

If v is the speed of the moving reference frame and if t is the time inside the reference frame and t’ the time outside:
AB = ct, AC = ct’ and BC = vt’
As AOB and AOC are right angle triangles, and if AO = y and OB = x :
(ct)² = x² + y² and (ct’)² = (x + vt’)² + y²
Therefore:
(ct)² - x² = (ct’)² - x² - 2xvt’ - (vt’)²
(ct)² = t’²(c² - v² - 2xv/t’)
t = t’ √(1 - v²/c² - 2xv/t’c²) ... (2)

Then imagine that the beam of light, as seen in the reference frame (AB), is pointing slightly behind the perpendicular.

If v is the speed of the moving reference frame and if t is the time inside the reference frame and t’ the time outside:
AB = ct, AC = ct’ and BC = vt’
As ACB is a right angle triangle:
(ct)² = (ct’)² + (vt’)²
t = t’ √(1 + v²/c²) ... (3)

This means that the proportion t : t’ varies according to the direction of the beam of light, with respect to that of the moving reference frame, and that the relativity of time depends on what is being shown.